A 4.9 g sample of a tribasic acid was dissolved in water to make 500 cm3 of solution. The pH of the solution was found to be 0.52. Calculate the relative molecular mass of the acid.

The given volume is 500 cm$^3$.

$$\text{Volume} = 500 \text{ cm}^3 \times \frac{1 \text{ L}}{1000 \text{ cm}^3} = 0.500 \text{ L}$$

The concentration of hydrogen ions is 0.3 M.

$$\text{Moles of H}^+ = \text{Concentration of H}^+ \times \text{Volume of solution}$$ $$\text{Moles of H}^+ = 0.3 \text{ mol/L} \times 0.500 \text{ L}$$ $$\text{Moles of H}^+ = 0.15 \text{ mol}$$

A tribasic acid produces 3 moles of hydrogen ions for every 1 mole of acid.
Let the acid be $\text{H}_3\text{A}$. The dissociation is:

$$\text{H}_3\text{A} \rightarrow 3\text{H}^+ + \text{A}^{3-}$$

Therefore, the moles of acid are one-third of the moles of hydrogen ions.

$$\text{Moles of acid} = \frac{\text{Moles of H}^+}{3}$$ $$\text{Moles of acid} = \frac{0.15 \text{ mol}}{3}$$ $$\text{Moles of acid} = 0.05 \text{ mol}$$

The mass of the acid is 4.9 g.

$$\text{Relative molecular mass} = \frac{\text{Mass of acid}}{\text{Moles of acid}}$$ $$\text{Relative molecular mass} = \frac{4.9 \text{ g}}{0.05 \text{ mol}}$$ $$\text{Relative molecular mass} = 98 \text{ g/mol}$$

The relative molecular mass of the acid is $\text{98}$.